EXCHANGE 


I  * 


An  Investigation  of  Comparative 

Deflections  of  Steel  Arch 

Ribs  with  Three,  Two 

and  No  Hinges 


AN  ABSTRACT  OF  A 
THESIS  PRESENTED 
TO  THE  FACULTY 
OF  THE  GRADUATE 
SCHOOL  FOR  THE 
DEGREE  OF  DOCTOR 
OF  PHILOSOPHY.  :  :  : 


By  PHOO  HWA  CHEN 
1917 


An   Investigation  of  Comparative 

Deflections  of  Steel   Arch 

Ribs  with  Three,  Two 

and  No  Hinges 


AN  ABSTRACT  OF  A 
THESIS  PRESENTED 
TO  THE  FACULTY 
OF  THE  GRADUATE 
SCHOOL  FOR  THE 
DEGREE  OF  DOCTOR 
OF  PHILOSOPHY.  :  :  : 


CORNELL  UNIVERSITY 


Reprinted  from  the  Cornell  Civil  Engineer 
vol.  XXVI,  pp.  184,  229,  Feb.,  Mar.,  1918 


By   PHOO   HWA   CHEN 
1917 


c> 


.*> 


385519 


DEFLECTION    OF   A   THREE-HINGED   ARCH 

PREFACE 

Early  in  boyhood,  before  the  writer  began  the  study  of  Civil  Engineer- 
ing, he  took  a  special  interest  in  arches  for  artistic  reasons.  While  in 
college,  his  natural  fondness  for  such  structures  led  him  to  become 
interested  in  the  advantages  and  disadvantages  of  one  type  over  another 
in  stiffness  and  economy.  The  object  of  this  thesis  is  to  investigate  the 
relative  stiffness  of  steel  arch  ribs  with  three,  two  and  no  hinges.  Since 
hingeless  arches  are  not  built  with  the  spandrel-braced  form  on  account 
of  the  difficulty  of  fixing  the  ends,  the  rib  type  is  chosen  for  this  inves- 
tigation. 

An  investigation  of  the  comparative  deflections  of  steel  arch  ribs  is  a 
very  complicated  problem.  The  design  of  a  three-hinged  arch  is  not 
affected  by  temperature  nor  by  rib-shortening.  The  effect  of  tempera- 
ture and  rib-shortening  on  arches  with  two  or  no  hinges  varies  for  differ- 
ent ratios  of  rise  to  span.  Finally,  different  designers  may  assume  differ- 
ent ranges  of  temperature  and  percentages  of  over-stress.  Realizing 
these  complications,  the  writer  paid  special  attention  to  finding  the 
easiest  methods  of  computation  for  the  benefit  of  future  investigators, 
rather  than  to  compute  only  the  value  for  deflection. 

The  general  process  of  finding  the  easiest  methods  of  computation 
used  in  this  thesis  is  to  analyze  general  equations  into  a  number  of  con- 
tributing factors ;  then  to  treat  each  factor  separately.  The  advantages 
of  this  process  are:  (i)  each  factor  may  have  a  very  simple  solution; 
(2)  it  gives  the  computer  a  clearer  conception  of  the  problem;  (3)  it 
offers  an  opportunity  to  study  the  relative  importance  of  different  con- 
tributing factors  and  to  neglect  some  of  the  factors;  (4)  it  may  suggest 
the  easiest  solution  by  omitting  certain  negligible  factors. 

The  special  features  of  this  thesis  are :  (i)  the  method  of  stress  com- 
putation for  the  three-hinged  arch;  (2)  the  method  of  computing 
deflections  for  the  three-hinged  arch;  (3)  the  method  of  computing 
deflections  due  to  axial  thrust  for  two  and  no  hinged  arches;  (4)  the 
assumption  of  moment  of  inertia  for  the  preliminary  design  of  the  hinge- 
less  arch;  (5)  the  method  of  stress  computation  for  the  hingeless  arch. 
Though  no  special  merit  is  claimed,  these  few  points  are  believed  to  be 
new. 

The  writer  wishes  to  express  his  appreciation  and  gratitude  for 
valuable  suggestions  and  encouragement  received  from  Prof.  H.  S. 
Jacoby,  chairman  of  the  committee. 


GENERAL    CONSIDERATIONS 

The  arches  chosen  for  this  investigation  give  some  advantage  to  two- 
hinged  and  no-hinged  arches  in  stiffness ;  because  the  lower  the  rise,  the 
greater  the  effect  of  rib-shortening  and  temperature  upon  cross-section 
area,  and  hence  the  smaller  the  deflection.  The  reason  for  choosing 
arches  with  a  low  rise  is  to  bring  out  the  good  points  of  the  three-hinged 
arch  even  under  unfavorable  conditions  and  a  desire  to  utilize  arches 
previously  designed.  The  arch  span  is  2  58  feet  long,  has  20  panels,  and  a 
rise  of  26  feet.  The  dead  load  was  assumed  to  be  59  kips  and  the  live 
load  i8>^  kips.  The  effective  depth  of  the  arch  rib  was  assumed  to  be 
five  feet  throughout,  and  the  unit-stress  15,000  pounds  per  square  inch. 
The  discovery  of  some  errors  led  the  writer  to  revise  the  designs  of  the 
three-hinged  and  no-hinged  arches,  and  thereby  to  find  some  new 
features  of  computation,  including  the  moment  table  for  the  three-hinged 
arch,  the  assumption  of  moments  of  inertia  for  the  preliminary  design  of 
tne  hingeless  arch,  and  the  effect  of  varying  the  end  cross-section. 

The  rise  and  fall  of  temperature  were  assumed  to  be  75°  Fahrenheit. 
Thirty  per  cent,  overs  tress  was  allowed  for  the  hingeless  arch,  but  no 
allowance  was  made  for  the  old  design  of  the  two-hinged  arch. 

THE    DESIGN 

The  method  of  stress  computation  used  in  this  design  is  different 
from  what  is  generally  employed.  The  writer  believes  that  the  new 
process  is  quicker  and  easier.  In  Chapter  V  of  Roofs  and  Bridges, 
Part  IV,  by  Professors  Merriman  and  Jacoby,  the  working  equations 
for  three-hinged  arches  are: 


ff-™  M-P(i-*)*-^        (a) 

2«  2« 

Equation  (a)  is  for  a  single  concentrated  load.  As  explained  in  the 
same  text,  each  leaf  of  the  arch  acts  both  as  a  simple  beam  and  a  strut. 
By  observing  this  simple  treatment,  much  labor  can  be  saved  in  com- 
puting the  bending  moment  at  different  sections  by  equation  (b) . 

M=M'  —  M"     (b) 

Where  M'  is  the  bending  moment  as  a  simple  beam,  and  M"  is  the  bend- 
ing moment  as  a  strut.  The  identity  of  equations  (a)  and  (b)  may  be 
proved  in  the  following  way : 

Vi'  =  P(i-2K);  VV  =  2PK  =  2F2;  M'  =  P  (i  -  2K)  oc 

TT  PKl 


Mf  -  M"  =  P  (i  -  2K)  x  - 


-  M"  = 


2h 


+  PKx 


M'  is  easily  computed  for  panel  loads.  If  M"  due  to  a  unit  load  at  the 
crown  hinge  is  known,  the  complete  moment  table  may  be  filled  in  by 
mere  inspection. 

For  a  unit  load  at  the  crown,  H  =  —  =  2.48077.     The  moment  at 


different  sections 

is  shown  in  table  i  . 

TABLE  i 

2hx 

Point 

y-.- 

i 

2-34 

2 

4.16 

3 

546 

4 

6.24 

5 

6.50 

6 

6.24 

7 

546 

8 

4.16 

9 

2.34 

10 

0.00 

5.805 

10.320 

13.545 
15.480 
16.125 
15.480 

13.545 
10.320 

5.805 

0.000 

For  any  other  position  of  the  load,  M"  will  be  a  simple  ratio  of  M"c. 
For  example,  when  k  =  0.25,  M"  =  o.$Mc";  when  k  =  0.2,  M"  = 
0.4  M"c.  The  bending  moment  diagram  for  a  simple  beam  is  a  triangle. 
If  the  moment  under  the  load  is  known,  the  rest  is  also  a  simple  ratio,  as 
shown  in  table  2,  because  the  panel  lengths  are  all  equal. 


Load  at 

point 

2 

4 
6 
8 


Load  at 
point 
Crown 
2 

4 
6 
8 


TABLE  2.     M'  TABLE 

Maximum  Coefficient  for  moment  at  section 

Moment  2468 


TABLE  3.     M"  TABLE 

Moment  at  section 
246 


0. 


0.8  M"* 


0.2M"4 

o.4M"4 
o.6M"4 
o.8M'/4 


0.2M" 


0.8M"' 


2                          4 

6                           8 

10.32              7.74 

5.16                       2.58 

20.64                   J5-48 

10.32                       5.16 

18.06                   20.22 

I5-48                      7-74 

15.48                   30.96 

20.64                    10.32 

12.90                   25.80 

25.80                    12.90 

10.32                   20.64 

30.96                    15.48 

7-74                   I5-48 

20.22                              18.06 

5.16                   10.32 

15.48                20.64 

2.58                     5.16 

7-74         1           10.32 

TABLE  4.     M'  TABLE 

Load  at  Moment  at  section 

point 
I 

2 

3 
4 
5 
6 

7 
8 


TABLE  5.     M"  TABLE 

Load  at  Moment  at  section 

point 

i 

2 

3 
4 
.5 
6 

7 
8 

9 
NOTE  —  The  zigzag  line  indicates  the  points  of  division. 

MAXIMUM    BENDING    MOMENTS 

Having  Tables  4  and  5  completed,  the  limiting  position  for  positive 
or  negative  moments  can  be  decided  at  once  by  comparing  M'  and  M"  '. 
This  is  indicated  by  zigzag  lines  in  tables  4  and  -5.  The  maximum 
moment  is  obtained  by  summing  up  the  moments  in  tables  4  and  5,  and 
multiplying  by  the  panel  load.  The  results  are  expressed  in  kip-feet. 

TABLE  6 

Section  Max.  L.  L.  Moment  D.  L.  Moment  Total 

2  1211.7  1-8  1213-5 

4  1747.2  1.8  1749-0 

6  1662.8  3.0  1665.7 

8  1033-5  3-0  1036.5 

FLANGE    AREA 

The  effective  depth  of  arch  rib  is  assumed  to  be  five  feet  and  the  unit- 
stress  15,000  pounds  per  square  inch. 

4 


2 

4 

6 

8 

1.32 

i-55 

i-55 

1.32 

2.06 

3.10 

3.10 

2.06 

3.10 

4.64 

4.64 

3.10 

4-13 

6.19 

6.19 

4.13 

5-16 

7-74 

7-74 

5.16 

6.19 

9.29 

9.29 

6.19 

7.22 

10.84 

10.84 

7.22 

8.26 

12.38     | 

12.38 

8.26 

9.29 

13-93 

13-93 

1          9.29 

Section 


0-2 


2-6 


6-8 


TABLE  7. 

FLANGE  AREA 

Max.  Moment 

Thrust 

Area  Required 

0.0 

2050 

68.0 

1213-5 

1862 

78.2 

1749.0 

1832 

84.4 

1665.7 

1770 

81.2 

1036.5 

1715 

71.0 

0.0 

1915 

64.0 

TABLE  8.     COMPOSITION   OF   FLANGES 

Composition 

Area 

I 

6Ls  6"  x  6"  x  ^ 

:"                  38.58  sq.  in. 

inch4 

3Pls.  I4"x  i/T 

2I.OO 

i  PI.  i6"x2<" 

12.00 

iPl.i6"Xiy 

7.00 

78.58 

155,496 

6  Ls  6"  x  6"  x  ^ 

*"                      38.58 

3  Pis.  I4"x>4" 

21.00 

i  PI.  i6"x^" 

12.00 

i  PI.  i6"Xly 

7.00 

I  PI.  I6"x^" 

6.00 

84.58 

173,162 

6  Ls  6"  x  6"  x  ^ 

i                38.58 

3  Pis.  14"  *1A" 

21.00 

i  PI.  i6"x^" 

12.00 

i  PI.  i6"x«^" 

10.00 

81.58 

163,652 

6  Ls  6"  x  6"  x  jS 

r"                38.58 

3  Pis.  i4"xK" 

21.00 

i  PI.  i6"x4<" 

12.00 

8-10 


71.58  136,912 

DEFLECTION 

There  is  hardly  any  method  in  existence  which  is  satisfactory  for 
computing  the  deflections  of  three-hinged  arch  ribs.  The  new  method 
to  be  applied  here  is  the  only  one  which  is  simple  and  suitable  for  com- 
puting the  deflection  of  various  points  along  the  axis  of  the  arch  rib. 

The  general  scheme  is  to  separate  the  total  deflection  into  its  com- 
ponent parts.  The  total  deflection  is  the  sum  of  the  deflections  in  the 
arch  rib  itself,  plus  that  due  to  the  crown  hinge  movement.  Again,  the 
deflection  of  the  arch  rib  itself  is  the  algebraic  sum  of  the  deflections  due 
to  positive  bending  moment  as  a  simple  beam,  and  due  to  negative 

5 


bending  moment  as  a  strut.  The  movement  of  the  crown  hinge  has 
three  contributing  factors:  the  deflection  due  to  negative  bending 
moment  in  both  leaves  of  the  rib;  the  deflection  due  to  positive 
bending  moment  in  either  leaf  of  the  arch  rib ;  and  the  deflection  due  to 
rib  shortening.  The  deflections  due  to  the  different  contributing  factors 
can  be  easily  obtained  either  graphically  or  analytically.  The  above 
statement  is  now  arranged  in  a  different  way  to  make  it  clearer. 

1.  Deflection  due  to  positive  bending  moment  as  a  simple  beam. 

2.  Deflection  due  to  negative  bending  moment  as  a  strut. 

3.  Influence  of  crown  movement  on  account  of  negative  bending 
moment  in  struts. 

4.  Influence  of  crown  movement  on  account  of  positive  bending 
moment  in  either  leaf  as  a  simple  beam. 

5.  Influence  of  crown  movement  on  account  of  rib  shortening. 


2O  Panels  g>  /<"?' 


Three-Hinged  Arch 

The  truth  of  the  above  method  may  be  proved  in  the  following  way. 
The  influence  of  the  work  of  shear  is  neglected.     The  deflection  is: 

ds 


JbMmds  CbTt 

EI  J      E 


A 


The  first  term  contains  the  first  four  contributing  factors.  The 
second  term  contains  the  fifth  contributing  factor.  Since  M  =  Mf  — M" 
and  m  =  m'  —  m",  the  first  term  becomes 

f  M™ds     =       f  (M'  -  M")  (m>  -  m")     *_ 
J      EI  J  El 

a  a 

rM'm'ds  _  rM"m'ds  _  (*  M'm'ds      C*  M'm'ds 
~ET~  ~J    ^T    ~J    ^T  +J      EI 

a  a  a  a 

The  first  three  terms  have  the  limits  a  and  c,  because  M'  and  m'  do  not 

6 


extend  to  the  other  half  of  the  arch.     By  mere  inspection,  the  four  terms 
are  the  first  four  contributing  factors.     The  expanded  equation  is, 

Ttds  . 


A 


J*M'm'ds_  £CM''m'ds      C° MWds      F*M"nf 
El        J     El         j      El         J    ^ 


EA 


Equation  (c)  may  be  changed  into  a  more  convenient  form  for  use. 
If/  equals  the  number  of  panel  points  from  the  nearest  support  to  where 
the  deflection  is  sought ;  and  g  equals  the  number  of  panel  points  from 
the  nearest  support  to  where  the  load  is  applied;  M"c  and  m"c  are  the 


Figure   6 


Figure  7. 


Deflection  Diagrams  with  Scales  Exaggerated 

bending  moments  due  to  a  load  unity  at  the  crown  instead  of  a  partial 
load  transferred  to  the  crown;   then  equation  (c)  becomes, 


+  fs 


rMc"mc"ds 
El 


(3) 


(4) 


(5) 


GRAPHICAL  REPRESENTATION    OF   CONTRIBUTING 
FACTORS   OF   DEFLECTION 

In  Fig.  5  the  left  leaf  of  the  original  arch  rib  is  a  e  c.     When  a  load  P 
is  applied,  the  leaf  deflects  as  indicated  by  ae"c'  in  exaggerated  scale. 

7 


The  ordinates  between  ae'c'  and  ae"cr  represent  the  deflections  as  a 
simple  beam;  and  this  is  called  the  contributing  factor  (i). 

When  a  load  P  is  applied  at  point  6,  0.6  p  is  transferred  to  the  crown 
hinge  which  causes  negative  bending  moment.  In  order  to  represent  the 
effect  of  positive  bending  moment  alone,  a  loadP'  =  0.6  p  is  supposed  to 
act  upwards  at  the  crown  hinge  to  counteract  the  effect  of  negative 
bending  moment.  On  account  of  factor  (i),  the  left  leaf  lengthens  and 
the  crown  hinge  moves  upwards  and  towards  the  right.  The  final 
position  of  the  arch  rib  is  ae"c'b.  The  ordinates  between  ae'c'b  and 
aecb  represent  the  effect  of  contributing  factor  (4).  The  ordinates 
between  aecb  and  ae"c'b  represent  the  combined  effect  of  (i)  and  (4)  as 
in  Fig.  5. 

When  a  partial  load  is  transferred  to  the  crown  hinge,  it  causes  nega- 
tive bending  moment  and  each  leaf  acts  as  a  strut.  On  account  of  the 
negative  bending  moment,  the  original  leaf  adc  in  Fig.  6  deflects  as 
ad'c'.  The  ordinates  between  ad'c'  and  ad"c'  represent  the  deflection 
due  to  factor  (2).  On  account  of  negative  bending  moment,  each  leaf 
shortens  and  the  crown  falls  to  its  new  position  c' .  The  ordinates 
between  adcb  and  ad"c'b  represent  the  deflection  due  to  factor  (3)  as 
shown  in  Fig.  6.  The  ordinates  between  adcb  and  ad'c'b  represent  the 
combined  effect  of  (2)  and  (3). 

The  effect  of  axial  thrust  is  simply  the  shortening  of  each  leaf  which 
causes  the  crown  to  fall  to  its  new  position.  In  Fig.  7  the  ordinates 
between  acb  and  ac'b  represent  the  deflection  due  to  factor  (5). 

As  may  be  readily  seen  from  the  figures,  factors  (3),  (4)  and  (5)  are 
due  to  crown  movement  and  can  be  combined  in  computing  the  deflec- 
tions at  various  points  for  any  given  loading,  when  the  movement  of  the 
crown  is  obtained. 

RECIPROCAL    DEFLECTIONS 

By  Maxwell's  law  of  reciprocal  deflections,  the  computation  is 
greatly  simplified.  For  example,  factor  (4)  is  obtained  directly  from 
factor  (3),  because  jM'M"ds!,El  and  /M'M'dslEI  are  identical. 
Furthermore,  Maxwell's  law  can  be  used  as  a  good  check  on  the  final 
results.  The  following  table  shows  the  scope  of  reciprocal  deflections. 
The  figures  i,  2,  3,  etc.,  indicate  reciprocal  deflections.  It  is  seen  that 
symmetrical  figures  on  both  sides  of  these  inclined  lines  are  equal. 

8 


LOAD  AT   SECTION 


Defl.  at. 

2 

4 

6 

8 

10 

12 
14 

16 

18 


II 


IV 


VI 


DEFLECTION   AS   A   SIMPLE   BEAM   BY   THE   METHOD    OF   ELASTIC   MOMENTS 

As  explained  on  page  223  of 
Bridge  Engineering  by  "Dr.  J.A.L. 
Waddell,  the  deflection  of  any 
point  of  a  simple  beam  with  respect 
to  its  supports  is  equal  to  the  mo- 
ment which  would  occur  if  the 
beam  were  conceived  to  be  simply 
supported  at  the  ends  and  loaded 
with  Mds/EI.  Factor  (i)  can  be 
easily  computed  by  this  method 
as  follows.  All  deflections  are 
to  be  computed  for  one  kip. 


ELASTIC   LOAD   FOR   LOAD   OF   ONE  KIP  AT  III 

(Mds/EI)  X  io6. 

Section  I  II          III  IV          V  VI  VII       VIII 

Elastic  load          4.40       8.25      11.60       8.60       8.15       6.65  5.10       3.68 


X 
2.03 


Section 
i 

2 

3 
4 
5 
6 

7 
8 

9 
10 


ds 


TABLE  9 
//looo       (ds/D  X  io3   (ds/EI)  X  10* 


A. 


13.72 

155.50 

.0082 

3.39 

78.58 

13.54 

164.33 

.0824 

3.17 

81.58 

13.39 

173.16 

.0773 

2.97 

84-58 

13.26 

173.16 

.0766 

2.95 

84.58 

13.15 

173.16 

.0759 

2.92 

84-58 

13.05 

168.41 

.0776 

2.99 

83.08 

12.98 

163.65 

.0793 

3.05 

81.58 

12.94 

150.28 

.0861 

3.32 

76.58 

12.92 

136.91 

.0945 

3.67 

71.58 

12.90 

136.91 

.0943 

3-66 

71.58 

DEFLECTION  OF  CROWN  DUE  TO  THRUST 

Thrust  is  equal  to  Hcosft  +  VsinQ.  For  a  load  unity  at  crown, 
H  =  2.48  and  V  =  0.5.  The  deflection  is  to  be  computed  as  it  affects 
all  points  along  the  arch  rib. 

The  result  is 


=   0.002 41 f  X  2   =  0.004834" 


2  A  E 


For  any  other  position  of  loading,  an  assumption  is  made  that  only  a 
portion  of  the  load  carried  up  to  the  crown  hinge  causes  crown  deflection 
due  to  rib-shortening.  This  assumption  is  justified  as  the  deflection  due 


—I HH — H 1 

Deflections  of  Three-Hinged  Arch 

to  rib-shortening  is  only  a  small  portion  of  the  total  and  it  does  not  make 
any  appreciable  error.  The  greatest  difference  occurs  when  the  load  is 
at  quarter  point.  The  difference  in  thrust  is  (V»  —  Vi)  sinft. 

10 


TABLE  10.     LOAD  AT  5 

Section  (7—70ri»e  d^^-  T2dsXio* 

2  A  E  2  AE 

o-i  .19  408  147 

1-2  .17  402  116 

2-3  -15  367  83 

3-4  -H  364  7i 

4-5  -12  36l  52 

469.0 

Total  deflection  due  to  thrust  =  0.00483400'' 

Deflection  due  to  (F2  —  Vi)  sinft  =  0.00000469" 

This  comparison  indicates  that  the  difference  is  a  negligible  quantity. 

Factors  (i)  and  (2)  are  computed  by  the  method  of  elastic  moments 
and  checked  by  graphical  methods.  In  order  to  describe  the  method, 
the  deflections  for  a  load  of  one  kip  at  point  IV  are  given  below.  Com- 
plete computations  are  filed  in  the  C.E.  Library  of  Cornell  University. 

1.  Compute  factors  (2),  (3)  and  (5)  for  a  load  unity  at  the  crown. 
These  values  are  used  for  any  position  of  a  load  with  a  corresponding 
ratio. 

2.  Compute  factor  (i)  for  load  at  IV. 
3  .     Compute  factor  (4)  for  load  at  IV. 

TABLE  n.     DEFLECTION  IN  1000TH  INCHES 

Point  (i) 

2  10.  6l 

4  16.91 

6  15-94 

8  945 
10 

12 
14 

16 

18 
(3),  (4)&(5)  =  10-849  —  0.4(14.566  —  4.834)  =  3.089 

(4)  (3)  (5) 

HORIZONTAL  DEFLECTION  DUE  TO  VERTICAL  LOAD 

An  identical  method  is  used  for  computing  horizontal  deflections 
except  that  m  stands  for  a  bending  moment  due  to  a  horizontal  load  of 
unity  and  /  for  the  thrust  due  to  the  same  loading.  The  general  equa- 
tion is, 


'w'ds       CbM"m'ds      CbM'm"ds  .     F*M*m*ds  .     C 

T  -j  -WT-]  —  +J  —  +  J 


II 


(2) 

(3),  (4),  and  (5) 

(i-5) 

2.728 

—  0.6178 

7.26 

—4-340 

—1.2356 

H.33 

—4.356 

—1.8534 

973 

2.760 

—2.4712 

4.22 

0.000 

—  3.0890 

—  3.09 

—  2.760 

—2.4712 

—  5-23 

—4.356 

—1.8534 

—  6.21 

—4.340 

—1.2356 

—  5.58 

—  2.728 

—  0.6178 

—  3.35 

rM'm'ds  _    C°M"m'ds  _    C°M'm"ds 
~ET      J    ~ET      J       El 


00 


oo 


(i)  (2)  (3)         (4)  (5) 

(4)  and  (5)  are  equal  to  zero,  because  the  summation  for  the  right  leaf  of 
the  arch  is  equal  and  opposite  to  the  summation  for  the  left  leaf.  The 
slight  variation  of  thrust  due  to  (Vz  —  Vi)  sin§  is  a  negligible  quantity. 
Deflections  as  a  simple  beam  and  as  a  strut,  are  computed  by  the  method 
of  elastic  moments. 


UHonz.       Deflection 
—Vert.   Load 1      _ 


onz.    Defltc  t  ion          [_ 


L v_    __    j 

Vert.cal          Deflect** 


VH- 


Honz.  Loa 


\Fiovre 


ion 
\ 


/ 


Fiqure  IS 


Deflections  of  Three-  Hinged  Arch;  Minimum  Deflections  shown  by  Dotted  Lines. 
HORIZONTAL   DEFLECTION   DUE    TO    HORIZONTAL    LOADING 

The  same  method  is  used  here  for  computing  the  horizontal  loading. 
In  this  case  all  terms  are  positive.  The  deflection  due  to  thrust  is  found 
to  be  very  small  and  that  it  is  beyond  the  degree  of  precision  required. 
The  general  formula  is, 


(I)  (2)  (3)  (4) 

Factors  (2)  and  (3)  have  a  reciprocal  relation.     When  one  is  known,  the 
other  can  be  obtained  by  a  simple  ratio.     Factor  (4)  needs  to  be  com- 


12 


puted  only  once,  as  the  rest  are  obtained  by  applying  a  simple  ratio. 
The  work  for  computing  the  deflection  is  thus  greatly  simplified. 

VERTICAL    DEFLECTION    DUE    TO    HORIZONTAL    LOADING 

Vertical  deflections  due  to  horizontal  loading  are  obtained  by 
Maxwell's  law  of  reciprocal  deflections,  because  the  vertical  deflection 
due  to  horizontal  loading  is  equal  to  the  horizontal  deflection  due  to 
vertical  loading. 


VERTICAL  DEFLECTION  OF  A  TWO-HINGED  ARCH 

The  general  formula  for  computing  deflection  is  fMmds  /El -\-fTtds  / 
A  E.  The  deflection  due  to  shear  is  to  be  neglected.  The  first  term  is 
the  deflection  due  to  bending  moment,  while  the  second  is  due  to 
shortening  of  the  arch  rib  on  account  of  axial  thrust.  Here  m  stands 
for  the  bending  moment  due  to  a  load  unity  applied  at  the  point  where 
the  deflection  is  sought.  Since  a  two-hinged  arch  is  a  combination  of  a 
simple  beam  and  strut,  the  bending  moment  at  different  sections  is 
equal  to  M'-M",  M'  being  the  bending  moment  when  acting  as  a  simple 
beam,  while  M"  is  the  bending  moment  when  acting  as  a  strut.  Simil- 
arly m  equals  m'-m".  Substituting  these  values  in  the  first  term,  it 
becomes  f(M'-M")  (m'-m")  ds/EI=fM'mfds/EI—fM"m'ds/EI—fM' 
m"ds  IEI+fM"m"ds  JET.  The  sum  of  the  last  two  terms  equals  zero, 
because  they  represent  the  vertical  deflection  due  to  a  movement  of 
the  end  hinges.  Since  the  hinges  do  not  move,  their  sum  must  be  zero. 
As  the  formula  shows,  the  first  term  of  the  expanded  formula  is  the 
deflection  when  acting  as  a  simple  beam ;  the  second  term  is  the  deflec- 
tion when  acting  as  a  strut  due  to  H;  the  third  term  is  the  deflection 
due  to  a  horizontal  movement  of  the  hinge  on  account  of  M' ;  the  fourth 
term  is  the  deflection  due  to  a  horizontal  movement  of  the  hinge  on 
account  of  M" ' .  Since  the  last  two  drop  out,  the  total  deflection  of  the 
arch  rib  due  to  flexure  equals  its  deflection  as  a  simple  beam  minus 
its  deflection  as  a  strut.  These  deflections  can  be  obtained  easily 
either  analytically  or  graphically.  The  complete  equation  becomes: 


Deflection 


ion  -  J, 


ds        C      ds 

rtf     f _i_     I    JY 

n  EJH  J     1AE 


(l) 


TABLE  12 


(2) 


(3) 


Section 

ds 

y 

/ 

0 

6.90 

2.50 

7-34 

i 

13.72 

4-94 

7-34 

2 

13-54 

9.36 

7.96 

3 

13-39 

13.26 

8-59 

4 

13.26 

16.64 

8.78 

5 

13-15 

I9-50 

8-97 

6 

13.05 

21.84 

8.97 

7 

12.98 

23.66 

8.99 

8 

12.94 

24.96 

8.80 

9 

12.92 

25-74 

8-59 

10 

12.90 

26.00 

8-59 

ds 

11 

yds/I 

y2ds/I 

( 

X94 

2-35 

5-88 

.87 

9-23 

45.60 

.70 

15.92 

149.00 

•56 

20.69 

274.10 

.51 

25.12 

418.10 

•47 

28.60 

557-50 

.46 

31.80 

694.00 

•44 

34-15 

808.00 

•47 

36.70 

915.00 

•51 

38.75 

999.00 

•50 

39-05 

1015.00 

0.018 

0.0300 

'  0.0370 

0.0395 

0.0395 

0.0300 

0.0550 

0.0700 

0.0760 

0.0752 

0.0375 

0.0700 

0.0935 

0.1050 

0.1050 

0.0405 

0.0760 

0.1045 

0.1220 

0.1250 

0.0395 

0.0750 

0.1050 

0.1250 

0.1327 

0.0354 

0.0765 

0.0950 

O.II50 

0.1250 

0.0285 

0.0550 

0.0780 

0.0955 

0.1050 

0.0200 

0.0390 

0.0550 

0.0680 

0.0752 

O.OIOO 

0.0200 

0.0290 

0.0355 

0.0395 

TABLE  13.     VERTICAL  DEFLECTION  AS  A  SIMPLE  BEAM.     (IN  INCHES) 

LOAD  AT  SECTION 

Point  II  IV  VI  VIII  X 

2 

4 

6 

8 

10 

12 

14 

16 

18 


TABLE  14.     VERTICAL  DEFLECTION  DUE  TO  H  =  UNITY.     (IN  INCHES) 

Point  Yds /I  Deflection 

1  9.23 

2  I5.92  O.O2IO 

3  20.69 

4  25.12  0.0500 

5  28.60 

6  31.80  0.0545 

7  34-15 

8  36.70  0.0637 

9  38.75 

10  39-°5  0.0670 


VERTICAL   DEFLECTION   DUE    TO    THRUST   FOR   A    VERTICAL    LOAD 

The  general  formula  for  the  deflection  due  to  thrust  is  fTtds  /AE 
in  which  T  is  the  thrust  due  to  the  applied  load  P',  while  t  is  the  thrust 
due  to  a  load  unity  P"  applied  where  the  deflection  is  sought.  Since 
the  deflection  due  to  axial  thrust  is  small  as  compared  with  that  of  the 
bending  moment,  and  the  axial  thrust  due  to  V  is  small  as  compared 
with  that  of  H,  the  effect  of  axial  thrust  due  to  V  may  be  neglected 
without  appreciable  error.  Let  H'  and  H"  be  the  horizontal  reactions 
due  to  P'  and  P"  respectively.  Then  the  formula  becomes: 

b 

Deflection  due  to  thrust  =  /  Hrcos§H"cos§ds  JAE 

a 
b 

=  /  H'H"cos2Qds/AE 

a 

With  a  systematic  arrangement,  the  computation  becomes  very  simple. 
As  seen  from  the  following  table,  there  are  many  reciprocal  products. 

15 


Point 

2 

4 

6 

8 

10 


II 

H*- 

H2* 


TABLE  15.     H'H" 
Load  at  Section 


IV 
H* 

H*H* 
H2* 

H*H* 

H'H" 


VI 

H6 


VIII 
H8 


H 


X 


H* 


H2 


There  is  a  very  simple  way  to  compute  the  deflection  due  to  an  axial 
thrust.     As  it  can  be  readily  observed  from  the  table,  the  deflection 


Deflections  of  a  Two-Hinged  Arch 

due  to  H  cos  6  is  proportional  to  H  in  both  directions.  Now  it  is  only 
necessary  to  compute  one  point  for  all  deflections  under  any  conditions 
of  loading.  This  is  the  deflection  of  the  crown  due  to  a  load  unity  at 
the  crown.  It  is  surprising  what  a  great  amount  of  labor  is  thus  saved. 
All  deflections  due  to  axial  thrust  therefore,  can  be  obtained  from  a 
simple  ratio  of  H .  As  stated  elsewhere  the  deflection  due  to  V  sin  0  may 
be  neglected.  If  more  accurate  results  are  desired,  the  effect  of  V 
should  be  computed  separately.  All  deflections  due  to  this  factor  are 
approximately  proportional  to  x. 

16 


The  method  of  combining  different  factors  is  illustrated  in  table  16, 
where  factor  (i)  stands  for  deflection  as  a  simple  beam,  factor  (2)  for 
deflection  as  a  strut,  and  factor  (3)  due  to  rib-shortening.  The  results 
are  plotted  for  panel  loads  of  one  kip.  Horizontal  deflections  are 
computed  by  identical  methods  except  that  P"  is  to  act  horizontally 
at  the  point  where  the  horizontal  deflection  is  sought. 


TABLE  1 6. 


Point 
2 

4 

6 

8 

10 

12 
14 

16 

18 


TOTAL  VERTICAL  DEFLECTION  DUE  TO  A  VERTICAL  LOAD 
AT  VI.     (IN  INCHES) 


Factor 

(i) 
•0375 
.0700 

•0935 
.1045 
.1050 
.0950 
.0780 
•°550 
.0285 


Factor 

~(2) 
.0329 
.O620 
•0845 
.0938 
.1040 
.0988 
.0845 
.O620 
.0329 


Factor 

(3) 
.0006 
.0012 
.0016 
.0020 
.0020 
.0020 
.0016 
.0012 
.0006 


Total 

.0052 

.0092 

.6106 

.0077 

— .0030 

— .0018 

— .0048 

—.0058 

—.0038 


DESIGN  OF  A  HINGELESS  ARCH 


Various  assumptions  of  the  moment  of  inertia  have  been  tried  for 
the  preliminary  design,  but  none  of  them  seems  to  be  adequate.  The 
preliminary  design  was  computed  by  assuming  I  to  vary  as  sec.  6.  The 
final  results  are  tabulated  as  follows : 


Load  at 
Section 

i 

2 

3 
4 
5 
6 

7 
8 

9 
10 


TABLE  17 
V*  H 


.9927 
.9720 
•9392 
.8960 

•8437 
.784o 
.7182 
.6480 

,5747 
.5000 


.0077 

.0818 

.0280 

.3013 

.0608 

.6026 

.1040 

•9523 

.1563 

1-3057 

.2160 

1.6405 

.2818 

1.9232 

.3520 

2.1427 

.4253 

2.2766 

.5000 

2.3250 

—121.3 
-  52.0 
— 28.9 

—17-3 
— 10.4 

—5-8 

—2.5 

o.oo 

1.9 

3.5 


F* 

10.04 

9-63 
9.18 
8.67 
8.09 

7-43 
6.67 
578 
4-73 
347 


When  an  allowance  is  made  for  30  per  cent,  over-stress  in  tempera- 
ture, the  following  flange  area  is  required. 


TABLE  18. 


Section 
o 

i 

2 

3 
4 

5 
6 

7 
8 

9 
10 


AREA  IN  SQUARE  INCHES 

A i  A2 

II9-5 
93-1 
75-9 
72.3 
65.0 

67-9 

72.8 
76.9 
78.4 
78-3 


77-3 


135-0 
106.5 

83-5 
74-7 
68.6 
59-2 
67.2 
72.8 
74-7 
75-8 
75-2 


Error 

—15-5 

13-4 

-7.6 

—2.4 

-3-6 

8.7 

5-6 

4.1 

3-7 
2-5 

2.1 


Al  is  computed  by  /  varies  as  sec.  0,  while,  A%  is  computed  by  true 
values  of  I.  The  error  is  far  too  great  for  the  preliminary  design. 
Evidently  the  assumption  I  varies  as  sec.  0  must  be  far  from  the  true 
value.  The  value  of  I  in  the  following  design  is  obtained  after  several 
trials.  The  moment  of  inertia  of  the  web  is  neglected. 


Section 
9-10 


TABLE  19.     COMPOSITION  OF  FLANGES 

Area  in-  I 

38.58 
23.60 

47-25 


Composition 
6—  Ls  6"  x  6"  X9"/i6 
3P1-  14  x  & 
3P1.  18  x  % 


8-9 


6-8 


4-6 


0-4 


2P1.  18 


6 — Ls  6  x  6  x  -ft 
3?1.  14  x  & 

3P1-  18  x  V8 


6 — Ls  6  x  6  x 
3P1-  14  x  A 
i  PI.  i8x  K 
iPl.  i8x& 


6— Ls  6  x  6  x 
3PL  14  x  A 

i  PI.  i8x  V* 


6 — Ls  6  x  6  x 
3P1.  14  x& 
i  PI.  18  x  ^ 
i  PI.  i8x  ^ 


27.00 

136.43 
38.58 
23-60 
47-25 

109.43 

38.58 
23.60 
I3-50 
10.13 

85.80 

38.58 

23-60 

7.88 

70.05 

38.58 

23.60 

7.88 

_6.75_ 

76.81 


316,900  in.4 


234,700  in. 


167,700  in.' 


125,850  inj 


143,400  in. 


18 


PROPOSED  METHOD  FOR  DESIGNING  A  HINGELESS  ARCH 
General  methods  for  designing  a  hingeless  arch  are  extremely 
laborious.  If  mistakes  are  made,  it  requires  a  long  time  to  correct 
them.  As  stated  by  Dr.  J.  A.  L.  Waddell  in  Bridge  Engineering,  page 
635,  "the  labor  involved  in  making  the  computations  is  excessive."  It 
seems  to  the  writer  that  the  following  interesting  method  could  be  used  to 
advantage.  A  hingeless  arch  is  a  two-hinged  arch  with  end  bending 
moments  added  to  keep  the  tangents  fixed.  Then  the  different  factors 
which  enter  the  computations  may  be  separated.  The  general  process 
is:  (i)  To  find  V  as  a  simple  beam;  (2)  to  find  H  as  a  two  hinge  arch; 


Moment  /nf/uente  Line, 
Tfof 

7fcc> 


(3)  to  find  61  and  02  due  to  the  applied  load;  (4)  to  find  0  due  to  H  unity; 
(5)  to  find  0i  and  02  due  to  M  =  unity  applied  at  one  end;  (6)  to  find 
H  and  V  due  to  M  =  unity  applied  at  one  end;  and  (7)  to  solve  the 
algebraic  equations  with  the  computed  data.  Factors  (i),  (2),  and  (6) 
are  simple,  but  the  rest  may  need  further  explanation. 

The  general  equation  for  0  is,  0  =  fMds  /EL  If  a  simple  supported 
beam  is  loaded,  the  point  of  maximum  deflection  will  be  the  point  of 
zero  shear  for  elastic  loads,  M ds  /El.  But  the  change  of  angle  at  each 
end  is  fMds  /El  from  the  end  to  the  point  of  maximum  deflection. 
Then  the  angles  61  and  02  are  simply  the  reactions  due  to  elastic  loads. 
Factors  (3),  (4)  and  (5),  therefore,  can  be  easily  computed.  It  is  to 
be  kept  in  mind,  when  M  =  unity  is  applied  at  one  end,  it  causes 
vertical  reactions  and  the  bending  moment  is  diminishing  from  unity 

19 


at  one  end  to  zero  at  the  other  end.     Moreover,  bending  moments  will 
cause  horizontal  reactions. 

The  algebraic  equations  are  all  simple ;  for  example,  if  61  and  02  are 
231  IE  and  42  /E  respectively,  and  M  =  unity  will  cause  61  =  11.02  /E 
and  62  =  6.2/E;  then 

11.02  MI-}-  6.2  MI  —  231 
11.02  M2  +  6.2  MI  =  42 
MI  and  M2  being  the  bending  moments  at  the  respective  supports. 

The  advantages  of  this  method  are:     (i)  some  factors  may  be  neg- 


Deflections  of  a  Hingeless  Arch 

lected;  (2)  if  mistakes  are  made,  they  can  be  easily  detected  and 
corrected.  The  writer  has  roughly  checked  one  point  with  this  method 
by  using  a  slide-rule  with  fairly  good  results.  Since  the  writer  has 
directed  his  attention  primarily  to  deflections,  he  is  not  able  to  spare 
the  time  to  develop  this  interesting  method.  Further  investigation 
is  necessary  in  order  to  make  this  method  a  useful  working  instrument. 


DEFLECTIONS  OF  A  HINGELESS  ARCH 

The  computations  of  deflections  for  a  hingeless  arch  are  quite  simple. 
There  are  two  main  factors:  bending  moment  and  axial  thrust.  Since 
a  hingeless  arch  is  fixed  at  both  ends,  the  arch  is  treated  as  a  cantilever 
for  computing  deflections  due  to  bending  moment.  The  plan  is  to  start 
from  one  end  and  to  use  the  other  as  a  check.  Since  their  relative 
position  is  fixed,  the  deflection  of  the  other  end  with  respect  to  the  first 
must  be  zero,  the  deflections  can  be  obtained  easily  either  analytically 
or  graphically. 

20 


The  deflection  due  to  axial  thrust  is  computed  in  exactly  the  same 
way  as  for  a  two-hinged  arch.  The  deflection  at  the  crown  is  first 
computed,  while  the  rest  are  obtained  by  means  of  a  simple  ratio  of  H. 

Horizontal  deflections  are  computed  by  similar  methods,  except 
that  the  load  unity  is  applied  horizontally  at  the  point  where  the  deflec- 
tion is  sought. 


TABLE   20. 


Point 

2 

4 

6 

8 

10 

12 
14 

1  6 

18 


VERTICAL   DEFLECTION    OF   A    HINGELESS   ARCH   FOR 
VERTICAL  LOAD  AT  X 


Factor 

(i) 

-  .52 

—  i  .08 

—  .60 
1.28 
2.60 

1.28 

-  .60 

—  i  .08 

—  .52 


Factor 

(2) 

40 

1.48 
2.79 
3.77 
4.13 

377 
2.79 

1.48 
.40 


Total 

-  .12 

.40 
2.19 

5.05 
6.73 

5.05 
2.19 

.40 
—  .12 


(A)      DISCUSSION  ON  THE  DESIGN 


The  new  method  of  stress  computation  for  a  three-hinged  arch 
seems  to  have  advantages  over  the  usual  methods  because  the  M'  and 
M"  (tables  4  and  5)  tables  can  be  filled  in  by  mere  inspection.  The 
point  of  division  can  be  readily  obtained  by  comparing  these  two  tables. 
This  method  can  be  applied  to  a  spandrel-braced  arch  as  well. 

Attention  has  been  called  to  the  assumption  of  moment  of  inertia 
for  the  preliminary  design  of  a  hingeless  arch.  The  writer  desired  to 
find  an  equation  for  the  /-curve  with  a  given  rise  but  on  account  of 
the  limited  time,  he  was  not  able  to  design  hingeless  arches  with  different 
rises.  The  best  that  the  writer  can  do  at  present  is  to  recommend  the 
/-curve  for  a  rise  equal  to  one-tenth  of  the  span.  He  hopes  that  some- 
one interested  in  this  problem  will  discover  the  necessary  equation. 
The  assumption  that  /  varies  as  sec  6  is  far  from  the  true  value.  For 
a  preliminary  design  of  hingeless  arches  with  one-tenth  rise,  the  assumed 
moment  of  inertia  should  be, 


Section 


Section 


10  (Crown) 
Ic 


c—  10% 


6 
c—  10% 


4 
Ic 


21 


3OO/OOOI 


isopoot 


Fig.  ^Q. 

Hinge/ess    Arch. 
I -Curve  For.1.  R/se. 


200000 


Before  choosing  the  final  flange  area,  y  =  fyds/I  +  fds/I  should  be 
investigated  to  see  how  much  it  differs  from  that  for  the  preliminary 
design.  This  is  important,  because  a  slight  variation  will  greatly 
affect  the  stresses  of  temperature  and  rib-shortening. 

Change  of  flang  area  does  not  materially  affect  H  nor  influence 
the  required  area  to  any 
great  extent,  except  near  the 
end  sections.  It  is  very 
interesting  to  note  that  too 
much  material  near  the  end 
section  may  reduce  the 
security  of  the  structure. 
In  designing  the  hingeless 
arch,  sections  i  and  o  were 
first  assumed  to  be  constant 
with  136  square  inches;  and 
the  final  required  area  for 
section  o  was  found  to  be 
142  square  inches.  When 
section  i  is  reduced  to  no 
square  inches,  while  the 
other  sections  remain  the 
same  as  before,  the  final 
area  for  section  o  becomes 
135  square  inches.  Too 
much  material  near  the  end 
section,  therefore,  is  undesir- 
able. The  explanation  is :  a  hingeless  arch  is  a  combination  of  a  canti- 
lever and  a  two-hinged  arch.  The  hinges  are  imaginary  and  movable. 
If  too  much  material  is  added  near  the  end  section,  it  increases  the  stiff- 
ness of  the  end  sections  unnecessarily,  and  in  turn  it  increases  the  action 
as  a  cantilever,  thus  requiring  a  larger  section  near  the  spring  line. 

(B)       DISCUSSION  ON  METHODS  OF  DEFLECTION  COMPUTATIONS 

The  method  of  computing  the  deflections  for  a  three-hinged  arch 
is  valuable  and  interesting.  The  computation  has  been  made  as  easy 
as  for  a  simple  beam.  Moreover,  this  new  method  presents  a  clear 
conception  of  the  different  contributing  factors. 

The  deflection  under  a  full  load  is  chiefly  due  to  axial  thrust.  It  is 
interesting  to  note  that  the  deflection  curve  for  a  three-hinged  arch 


22 


under  full  load  is  composed  of  two  straight  lines  (see  Fig.  33). 
This  clearly  shows  that  the  deflection  due  to  axial  thrust  is  proportional 
to  x  which  in  turn  is  proportional  to  H,  and  V.  For  arches  with  two 
and  no  hinges,  the  deflections  under  full  load  are  very  nearly  propor- 
tional to  H  (see  Fig.  33). 

The  method  of  computing  deflections  due  to  axial  thrust  for  arches 
with  two  and  no  hinges  is  useful  and  simple.  It  is  only  necessary  to 
compute  for  one  point  to  obtain  all  deflections  under  all  kinds  of  loading. 


K         ^---~^  |  |     A 


M\  —  —  -      y/  .for  Vtrticol  Loattar  o 

Xl4—  U      -^_ 


Comparison  of  Deflections  for  Hingeless,  Two-Hinged  and  Three-Hinged  Arches 

23 


(c)       DISCUSSION  ON  DEFLECTION 

The  relative  deflections  for  load  at  II,  IV,  VI,  VIII  and  X  are  plotted 
in  diagrams  26  to  54  inclusive;  and  comparisons  are  made  for  horizontal 
and  vertical  deflections  due  to  horizontal  and  vertical  loads.  Finally 
comparisons  are  made  for  loads  giving  maximum  and  minimum  deflec- 
tions and  for  full  loads. 

So  far  as  the  deflection  of  the  crown  is  concerned,  a  two-hinged  arch 
is  stiff er  than  a  three-hinged  arch  for  full  and  partial  loads.  At  the 
quarter  points,  however,  the  reverse  is  true.  The  hingless  arch  is  the 
stiff est  of  all  under  all  conditions. 


Morn    J)€tle<,ttOn$  for  Hc^t  Load 


Comparison  of  Deflections  for  Hingeless,  Two-Hinged  and  Three-Hinged  Arches 

When  the  arches  are  under  horizontal  loads,  the  relative  deflections 
of  arches  with  two  and  three  hinges  remain  the  same  as  before  stated. 
While  under  full  loads,  there  is  practically  no  difference  in  deflection 
for  these  two  types  of  arches.  The  deflection  for  a  hingeless  arch  due 
to  horizontal  loads  is  a  negligible  quantity;  while  for  arches  with  two 
and  three  hinges,  the  horizontal  deflection  due  to  horizontal  loads  is 
about  15  per  cent,  of  Vertical  Deflection  at  the  quarter  points. 

The  hingeless  arch  is  the  stiffest  both  under  horizontal  and  vertical 
loads.  It  has  very  small  deflections  due  to  horizontal  loads,  and  hence 
the  vibration  due  to  a  fast  railroad  train  is  greatly  reduced.  In  the 
writer's  opinion,  with  a  favorable  location,  the  hingeless  arch  would  be 
the  first  choice  if  the  greatest  stiffness  is  required.  Under  ordinary 
conditions,  a  three-hinged  arch  is  preferable. 


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